Integral log Gamma 0 to 1/2

In this post we will proof the following two integrals related to the derivative of the Riemann Zeta function at 2


\int_0^{1/2}\ln\left(\Gamma(x)\right)dx=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi


\int_0^\infty \frac{x \ln x}{e^{\pi x}-1}\,dx=\frac16+\frac{\ln 2}{6}-2 \ln A

Recall Kummer´s fourier expansion for LogGamma   0<x<1


\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right)(1)

Integrating (1) from 0 to x


\int_0^x\ln\left(\Gamma(u)\right)du=\frac{x \ln\left(2 \pi\right)}{2}+\frac{1}{4 \pi}\sum_{k=1}^{\infty}\frac{\sin\left(2 \pi k x \right)}{k^2} +\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\cos \left(2 \pi k x \right)}{k^2}+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right)}{k^2}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right) \cos\left(2 \pi k x \right)}{k^2}(2)

Letting x=\frac12 in (2)


\begin{aligned} \int_0^{1/2}\ln\left(\Gamma(x)\right)dx&=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{2 \pi^2}\zeta(2)+\frac{\gamma}{2 \pi^2}\eta(2)+\frac{\ln(2\pi)}{2 \pi^2}\zeta(2)+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k\right)}{k^2}+\frac{\ln \left( 2 \pi \right)}{2 \pi^2}\eta(2)-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k \right) \left(-1 \right)^{k}}{k^2}\\ &=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{12 }+\frac{\gamma}{24}+\frac{\ln(2\pi)}{12 }-\frac{1}{2 \pi^2}\zeta^{\prime}(2)+\frac{\ln \left( 2 \pi \right)}{24}-\frac{1}{2 \pi^2}\eta^{\prime}(2)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{1}{2 \pi^2} \cdot \frac{\pi^2}{6}\left(\ln(2 \pi)+\gamma-12 \ln A \right)-\frac{1}{2 \pi^2} \left(\frac{\pi^2 \ln(2 )}{12} +\frac{\pi^2}{12}\left(\ln(2 \pi)+\gamma-12 \ln A\right)\right)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{\ln(2 \pi)}{12}-\frac{\gamma}{12}+\ln A-\frac{\ln (2)}{24}-\frac{\ln(2 \pi)}{24}-\frac{\gamma}{24}+\frac{\ln A}{2}\\ &=\left(\frac{1}{8}-\frac{1}{12}-\frac{1}{24} \right)\gamma+\left(\frac38-\frac{1}{12}-\frac{1}{24} \right)\ln(2 \pi)-\frac{\ln(2)}{24}+\ln A+\frac{\ln A}{2}\\ &=\frac{3}{2}\ln A+ \frac{6}{24}\ln (2 \pi)-\frac{\ln (2)}{24}\\ &=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi \qquad \blacksquare \end{aligned}

Where we used that 

\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln 2 \pi + \gamma -12 \ln A \right)

Proved here, and that

\eta^{\prime}(s)=2^{1-s}\ln (2) \zeta(s)+(1-2^{1-s})\zeta^{\prime}(s)

For the second integral, lets first evalute the following integral


                                              \begin{aligned} I(s)&=\int_0^\infty \frac{x^{s-1}}{e^{cx}-1}\,dx\\ &=\frac{1}{c^s}\int_0^\infty \frac{x^{s-1}}{e^{x}-1}\,dx\\ &=\frac{1}{c^s}\int_0^\infty \frac{e^{-x}x^{s-1}}{1-e^{-x}}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \, \int_0^\infty x^{s-1}e^{-kx}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\ &=\frac{\Gamma(s)\zeta(s)}{c^s} \end{aligned}

Then, differentiating both sides w.r. to s


\begin{aligned} \int_0^\infty \frac{x^{s-1} \ln x}{e^{cx}-1}\,dx&=\frac{d}{ds}\left(\frac{\Gamma(s)\zeta(s)}{c^s} \right)\\ &=\frac{c^s\left(\Gamma^{\prime}(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-c^s\ln c \left( \Gamma(s)\zeta(s) \right)}{(c^s)^2}\\ &=\frac{\left(\Gamma(s)\psi(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-\ln c \left( \Gamma(s)\zeta(s) \right)}{c^s}\\ \end{aligned}

Setting s=2


\begin{aligned} \int_0^\infty \frac{x \ln x}{e^{cx}-1}\,dx&=\frac{\left(\Gamma(2)\psi(2)\zeta(2)+\Gamma(2)\zeta^{\prime}(2) \right)-\ln c \left( \Gamma(2)\zeta(2) \right)}{c^2}\\ &=\frac{\left((1-\gamma)\zeta(2)+\zeta(2)\left(\ln 2\pi+\gamma-12 \ln A\right) \right)-\zeta(2)\ln c }{c^2}\\ &=\zeta(2)\left(\frac{1-\ln c+\ln 2\pi-12 \ln A }{c^2}\right)\\ \end{aligned}

Setting c=\pi

\int_0^\infty \frac{x \ln x}{e^{\pi x}-1}\,dx=\frac16+\frac{\ln 2}{6}-2 \ln A \qquad \blacksquare


Where we again made use of the fact that


\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln 2 \pi + \gamma -12 \ln A \right)


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